# Thread: Stumped on two circular trig problems - help please!

1. ## Stumped on two circular trig problems - help please!

I've got two problems here that really have me stumped. The first one, I'm getting two answers algebraically that I'm not getting graphically. The second one, I'm getting a handful of answers graphically that I'm not getting algebraically. Can someone help me figure out what I'm doing wrong?

2. Originally Posted by ChrisEffinSmith
I've got two problems here that really have me stumped. The first one, I'm getting two answers algebraically that I'm not getting graphically. The second one, I'm getting a handful of answers graphically that I'm not getting algebraically. Can someone help me figure out what I'm doing wrong?
first one: cos(x) = -2 does not have any solutions because $\displaystyle -1 \leq cos(x) \leq 1$. Otherwise your working looks good and corresponds to (0,2) on the graph

second one: you're using the graph of cos(5x) which makes the points be squashed closer together.

3. Originally Posted by e^(i*pi)
first one: cos(x) = -2 does not have any solutions because $\displaystyle -1 \leq cos(x) \leq 1$. Otherwise your working looks good and corresponds to (0,2) on the graph

second one: you're using the graph of cos(5x) which makes the points be squashed closer together.
So for the first one it's acceptable to simply say:
$\displaystyle cos(x) \neq -2$ ?

And for the second problem, I kind've understand what you're saying about 5x being a transformation of x. So because my algebraic solution is for x and not 5x, I'm stretching out what is otherwise compressed on the graph? Is it possible to obtain those other values algebraically? Thanks!

4. Originally Posted by ChrisEffinSmith
So for the first one it's acceptable to simply say:
$\displaystyle cos(x) \neq -2$ ?

And for the second problem, I kind've understand what you're saying about 5x being a transformation of x. So because my algebraic solution is for x and not 5x, I'm stretching out what is otherwise compressed on the graph? Is it possible to obtain those other values algebraically? Thanks!
It would be better to say that there are no values of x that satisfy the equation cos(x)=-2 or something along those lines because cos(x)=-2 is a solution to the quadratic but no value of x satisfies it.

For the second problem you can either extend your limits by 5 times from 0 to 10pi and do what you did before to find more solutions or you can do 2pi/5 - your solution. Personally, I think the first way is easier